two electrons are released towards each other with equal velocities of 10 to the power 6 meter second-1. what will be the closest approach between them
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1
Answer:
2.53 Å
Explanation:
Total kinetic energy of electrons
K.E
=
2
×
1
2
mv
2
Potential energy of electrons
U
=
ke
2
r
When they approached closest distance (
r
) their total kinetic energy is converted into potential energy.
K.E = U
2
×
1
2
mv
2
=
ke
2
r
r
=
ke
2
mv
2
r
=
9
×
10
9
Nm
2
/
C
2
×
(
1.6
×
10
−
19
C
)
2
9.1
×
10
−
31
kg
×
(
10
6
m/s
)
2
=
2.53
×
10
−
10
m
=
2.53 Å
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